Integrand size = 27, antiderivative size = 178 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {(i a+b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}-\frac {(i a-b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{3 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f} \]
(I*a+b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f/(c-I*d)^(1/2)-(I *a-b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f/(c+I*d)^(1/2)-4/3* b^2*(-4*a*d+b*c)*(c+d*tan(f*x+e))^(1/2)/d^2/f+2/3*b^2*(c+d*tan(f*x+e))^(1/ 2)*(a+b*tan(f*x+e))/d/f
Time = 1.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 \left (-\frac {3 i (a-i b)^3 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 \sqrt {c-i d}}+\frac {3 i (a+i b)^3 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 \sqrt {c+i d}}+\frac {2 b^2 (-b c+4 a d) \sqrt {c+d \tan (e+f x)}}{d}+b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}\right )}{3 d f} \]
(2*((((-3*I)/2)*(a - I*b)^3*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I* d]])/Sqrt[c - I*d] + (((3*I)/2)*(a + I*b)^3*d*ArcTanh[Sqrt[c + d*Tan[e + f *x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + (2*b^2*(-(b*c) + 4*a*d)*Sqrt[c + d*Ta n[e + f*x]])/d + b^2*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]))/(3*d* f)
Time = 0.90 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4049, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle \frac {2 \int \frac {3 d a^3-2 b^2 (b c-4 a d) \tan ^2(e+f x)-b^2 (2 b c+a d)+3 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 d a^3-2 b^2 (b c-4 a d) \tan ^2(e+f x)-b^2 (2 b c+a d)+3 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 d a^3-2 b^2 (b c-4 a d) \tan (e+f x)^2-b^2 (2 b c+a d)+3 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {\int \frac {3 a \left (a^2-3 b^2\right ) d+3 b \left (3 a^2-b^2\right ) \tan (e+f x) d}{\sqrt {c+d \tan (e+f x)}}dx-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a \left (a^2-3 b^2\right ) d+3 b \left (3 a^2-b^2\right ) \tan (e+f x) d}{\sqrt {c+d \tan (e+f x)}}dx-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}+\frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3}{2} d (a-i b)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d (a+i b)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3}{2} d (a-i b)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d (a+i b)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3 i d (a-i b)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {3 i d (a+i b)^3 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {3 i d (a-i b)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {3 i d (a+i b)^3 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3 (a-i b)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {3 (a+i b)^3 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3 d (a-i b)^3 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {3 d (a+i b)^3 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}-\frac {4 b^2 (b c-4 a d) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}\) |
(2*b^2*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + ((3*(a - I *b)^3*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (3*(a + I* b)^3*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (4*b^2*(b*c - 4*a*d)*Sqrt[c + d*Tan[e + f*x]])/(d*f))/(3*d)
3.13.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3850\) vs. \(2(152)=304\).
Time = 1.02 (sec) , antiderivative size = 3851, normalized size of antiderivative = 21.63
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3851\) |
derivativedivides | \(\text {Expression too large to display}\) | \(8262\) |
default | \(\text {Expression too large to display}\) | \(8262\) |
a^3*(1/4/f/d/(c^2+d^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^ 2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+1/4 /f*d/(c^2+d^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2) +2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/4/f/d/(c^2+d^ 2)^(3/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^ (1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-1/4/f*d/(c^2+d^2) ^(3/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1 /2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-1/f/d/(c^2+d^2)^(1/2) /(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^ 2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-1/f*d/(c^2+d^2)^(1 /2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2 +d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+1/f/d/(c^2+d^2)^(3/ 2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+ d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^4+3/f*d/(c^2+d^2)^ (3/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c ^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+2/f*d^3/(c^2+ d^2)^(3/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+ (2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-1/4/f/d/(c^2 +d^2)*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/ 2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/f*d/(c^2+d^2)...
Leaf count of result is larger than twice the leaf count of optimal. 3928 vs. \(2 (146) = 292\).
Time = 0.51 (sec) , antiderivative size = 3928, normalized size of antiderivative = 22.07 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
-1/6*(3*d^2*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 1 18*a^6*b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^2 - 4*(3*a^11*b - 55*a^9*b^3 + 198 *a^7*b^5 - 198*a^5*b^7 + 55*a^3*b^9 - 3*a*b^11)*c*d + (a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)*c + 2*(3 *a^5*b - 10*a^3*b^3 + 3*a*b^5)*d)/((c^2 + d^2)*f^2))*log((2*(3*a^11*b - a^ 9*b^3 - 18*a^7*b^5 - 18*a^5*b^7 - a^3*b^9 + 3*a*b^11)*c - (a^12 - 12*a^10* b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^12)*d)*sqrt(d*tan(f*x + e) + c) + (((a^3 - 3*a*b^2)*c^3 + (3*a^2*b - b^3)*c^2*d + (a^3 - 3*a*b^2)*c* d^2 + (3*a^2*b - b^3)*d^3)*f^3*sqrt(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6 *b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^2 - 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b ^5 - 198*a^5*b^7 + 55*a^3*b^9 - 3*a*b^11)*c*d + (a^12 - 30*a^10*b^2 + 255* a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)*d^2)/((c^4 + 2*c ^2*d^2 + d^4)*f^4)) + (2*(9*a^7*b^2 - 33*a^5*b^4 + 19*a^3*b^6 - 3*a*b^8)*c ^2 - (9*a^8*b - 84*a^6*b^3 + 126*a^4*b^5 - 36*a^2*b^7 + b^9)*c*d + (a^9 - 18*a^7*b^2 + 60*a^5*b^4 - 46*a^3*b^6 + 3*a*b^8)*d^2)*f)*sqrt(-((c^2 + d^2) *f^2*sqrt(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6*b^6 - 60*a^4*b^8 + 9*a^2* b^10)*c^2 - 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 55*a^3* b^9 - 3*a*b^11)*c*d + (a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 25 5*a^4*b^8 - 30*a^2*b^10 + b^12)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (...
\[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
Time = 12.03 (sec) , antiderivative size = 3017, normalized size of antiderivative = 16.95 \[ \int \frac {(a+b \tan (e+f x))^3}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
atan(((((8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3*f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b* c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20* a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^2 - b^6*d^2 + 15*a^2*b^4*d^2 - 15*a^4*b^2*d^2))/f^2)*(( 6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*1 5i)/(4*(c*f^2*1i - d*f^2)))^(1/2)*1i - (((8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3* f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3* b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2))*((6*a*b^5 + 6*a^5*b - a^ 6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f ^2)))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^2 - b^6*d^2 + 15*a^2*b ^4*d^2 - 15*a^4*b^2*d^2))/f^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2 *b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2)*1i)/((( (8*(4*a^3*d^3*f^2 - 12*a*b^2*d^3*f^2 + 4*b^3*c*d^2*f^2 - 12*a^2*b*c*d^2*f^ 2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)) )^(1/2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*(c*f^2*1i - d*f^2)))^(1/2) - (16*(c + d*tan(e + f*x))...